[R] unbalanced design

[Peter Alspach]

Damián

I asked a similar question a few months ago (3 August 2004):

> temp.aov <- aov(S~rep+trt1*trt2*trt3, data=dummy.data)
> model.tables(temp.aov, type='mean', se=T)
>
> Returns the means, but states "Design is unbalanced - use se.contrasts
> for se's" which is a little surprising since the design is balanced. 

To which Prof Ripley replied: If you used the default treatment contrasts, it is not.  Try Helmert
contrasts with aov().

If I recall correctly, following Prof Ripley's suggestion led aov() to accept the design was balanced, but model.tables() still did not (but that could have been my error).  However, se.contrast() worked.

Cheers ........

Peter Alspach



>>> Damián Cirelli <damian.cirelli at maine.edu> 02/12/04 09:50:13 >>>
Hi all,
I'm new to R and have the following problem:
I have a 2 factor design (a has 2 levels, b has 3 levels). I have an
object kidney.aov which is an aov(y ~ a*b), and when I ask for
model.tables(kidney.avo, se=T) I get the following message along with
the table of effects:

Design is unbalanced - use se.contrast() for se's

but the design is NOT unbalanced... each fator level combination has the
same n

I' d appreciate any help.
Thanks.

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